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Math Help - locus of orthocenters

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    locus of orthocenters

    Determine the locus of the point O of intersection of the altitudes (orthocenter) of a triangle ABC, if the locus of the vertex A is a line parallel to BC.
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    Quote Originally Posted by scipa View Post
    Determine the locus of the point O of intersection of the altitudes (orthocenter) of a triangle ABC, if the locus of the vertex A is a line parallel to BC.
    The locus of the orthocenter is a parabola passing through B and C.

    Let \alpha denote the interior angle at A, then

    1. the parabola cuts the parallel to BC through A in 2 points if \alpha > 90^\circ;

    2. the parallel to BC through A is tangent to the parabola if \alpha = 90^\circ;

    3. the vertex of the parabola is located between the parallel to Bc and the line BC if \alpha < 90^\circ;
    Attached Thumbnails Attached Thumbnails locus of orthocenters-locus_parab.gif  
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  3. #3
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    Quote Originally Posted by scipa View Post
    Determine the locus of the point O of intersection of the altitudes (orthocenter) of a triangle ABC, if the locus of the vertex A is a line parallel to BC.
    Second attempt:

    Place the base BC of the triangle on the x-axis. Let d denote the distance between the 2 parallels. B(0,0) and C(c,0) and A(x,d).

    The slope of AC is m_{AC}=-\frac d{c-x}. Then the height through B (perpendicular to AC) has the slope:

    m_{h_B}=\frac{c-x}d

    Then the orthocenter is H \left( x,\frac{(c-x)x}d\right)

    That means all orthocenters lie on the curve of

    y=-\frac1dx^2+\frac cdx

    I've choosen d = 5, C(8,0)
    Attached Thumbnails Attached Thumbnails locus of orthocenters-hoehenparabel.jpg  
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